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What Is The Population Mean Of The Times To Repair?

Written By Saturday, April 16, 2022 Add Comment Edit

The uniform distribution is a continuous probability distribution and is concerned with events that are every bit likely to occur. When working out problems that have a uniform distribution, be careful to notation if the information is inclusive or sectional.

Example i

The data in the table below are 55 smiling times, in seconds, of an 8-week-sometime babe.

ten.4 nineteen.half dozen xviii.viii 13.ix 17.eight 16.viii 21.6 17.9 12.5 eleven.1 four.9
12.8 14.8 22.8 twenty.0 xv.9 16.three 13.iv 17.i 14.five 19.0 22.8
i.3 0.7 8.9 11.9 ten.nine vii.3 five.nine 3.vii 17.9 nineteen.2 9.8
5.8 6.nine 2.6 v.eight 21.7 11.8 iii.4 ii.1 4.v 6.iii x.7
eight.nine 9.4 9.4 7.six ten.0 iii.3 six.7 vii.8 11.half-dozen 13.8 xviii.vi

The sample mean = 11.49 and the sample standard deviation = 6.23.

We will presume that the smiling times, in seconds, follow a compatible distribution between zero and 23 seconds, inclusive. This means that whatsoever smiling time from zero to and including 23 seconds is every bit likely. The histogram that could be synthetic from the sample is an empirical distribution that closely matches the theoretical uniform distribution.

LetX = length, in seconds, of an viii-week-old baby'due south smile.

The notation for the uniform distribution isX ~ U(a, b) where a = the everyman value of x and b = the highest value of ten.

The probability density function is [latex]\displaystyle{f{{({x})}}}=\frac{{i}}{{{b}-{a}}}\\[/latex] for a ≤ 10 ≤ b.

For this example,X ~ U(0, 23) and [latex]\displaystyle{f{{({x})}}}=\frac{{ane}}{{{23}-{0}}}\\[/latex] for 0 ≤ X ≤ 23.

Formulas for the theoretical mean and standard deviation are [latex]\displaystyle{\mu}=\frac{{{a}+{b}}}{{2}}{\quad\text{and}\quad}{\sigma}=\sqrt{{\frac{{{({b}-{a})}^{{2}}}}{{12}}}}\\[/latex]

For this trouble, the theoretical mean and standard difference are [latex]\displaystyle{\mu}=\frac{{{0}+{23}}}{{2}}={11.50} \text{ seconds}{\quad\text{and}\quad}{\sigma}=\sqrt{{\frac{{{({23}-{0})}^{{ii}}}}{{12}}}}={6.64} \text{ seconds}\\[/latex]

Find that the theoretical mean and standard departure are close to the sample mean and standard deviation in this example.

Try It

The data that follow are the number of passengers on 35 unlike charter line-fishing boats. The sample mean = 7.ix and the sample standard divergence = 4.33. The data follow a uniform distribution where all values betwixt and including zero and 14 are equally likely. State the values ofa and b. Write the distribution in proper notation, and calculate the theoretical hateful and standard departure.

one 12 iv 10 4 14 11
7 xi 4 13 2 4 half dozen
3 10 0 12 6 9 10
v 13 four ten 14 12 11
vi 10 11 0 11 13 2

a is aught; b is 14; Ten ~ U (0, 14); μ = seven passengers; σ = 4.04 passengers

Instance 2

  1. Refer to Case 1 What is the probability that a randomly chosen eight-week-old baby smiles between two and xviii seconds?
  2. Observe the 90th percentile for an eight-calendar week-sometime baby'due south grin time.
  3. Notice the probability that a random eight-week-former baby smiles more than 12 seconds knowing that the baby smiles more than eight seconds.

Solution

  1. Notice P(two < 10 < 18).
    [latex]\displaystyle{P}{({ii}{<}{10}{<}{18})}={(\text{base})}{(\text{height})}={({18}-{two})}{(\frac{{1}}{{23}})}={(\frac{{16}}{{23}})}.\\[/latex]This graph shows a uniform distribution. The horizontal axis ranges from 0 to 15. The distribution is modeled by a rectangle extending from x = 0 to x = 15. A region from x = 2 to x = 18 is shaded inside the rectangle.
  2. 90 percentage of the smile times fall below the 90th percentile, k, so P(x <m) = 0.90
    [latex]P(x<k)=0.90\\[/latex]
    [latex](\text{base of operations})(\text{height})=0.90\\[/latex]
    [latex]\displaystyle(g-0)(\frac{1}{23})=0.ninety\\[/latex]
    [latex]yard=(23)(0.90)=twenty.7\\[/latex]
    This shows the graph of the function f(x) = 1/xv. A horiztonal line ranges from the point (0, 1/fifteen) to the indicate (15, 1/15). A vertical line extends from the x-axis to the cease of the line at bespeak (15, 1/xv) creating a rectangle. A region is shaded within the rectangle from x = 0 to x = k. The shaded area represents P(10 < k) = 0.90.
  3. This probability question is a conditional. You are asked to discover the probability that an 8-week-sometime infant smiles more than than 12 seconds when you already know the infant has smiled for more than eight seconds.ObserveP(x > 12|x > 8) There are two ways to do the problem.
    1. For the first way, use the fact that this is a provisional and changes the sample space. The graph illustrates the new sample space. You already know the infant smiled more than eight seconds.Write a newf(x):[latex]\displaystyle{f{{({x})}}}=\frac{{1}}{{{23}-{8}}}=\frac{{1}}{{xv}}\\[/latex]for 8 < ten < 23
      f(X)=1/15 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1/15 on the y-axis, a vertical upward line from points 8 and 23 on the x-axis, and the x-axis. A shaded region from points 12-23 occurs within this area.
    1. For the 2d way, use the conditional formula (shown below) with the original distribution X ~ U (0, 23):For this problem,A is (10 > 12) and B is (x > eight).
      This shows the graph of the function f(x) = 1/23. A horiztonal line ranges from the point (0, 1/23) to the point (23, 1/23). A vertical line extends from the x-axis to the end of the line at point (23, 1/23) creating a rectangle. Vertical lines extend from the horizontal axis to the graph at x = 8 and x = 12.

Endeavour It

A distribution is given asX ~ U (0, 20). What is P(2 < x < eighteen)? Discover the 90th percentile.

P(2 < x < xviii) = 0.8; 90th percentile = 18

Example 3

The amount of time, in minutes, that a person must wait for a motorcoach is uniformly distributed between cypher and 15 minutes, inclusive.

  1. What is the probability that a person waits fewer than 12.five minutes?
  2. On the average, how long must a person look? Discover the mean, μ, and the standard deviation, σ.
  3. Ninety percent of the time, the time a person must wait falls below what value? This asks for the 90th percentile.

Solution

  1. Permit X = the number of minutes a person must wait for a bus. a = 0 and b = xv. X~ U(0, 15). Write the probability density function. [latex]\displaystyle{f{{({x})}}}=\frac{{one}}{{{xv}-{0}}}=\frac{{1}}{{15}}\\[/latex] for 0 ≤x ≤ xv.DetectP (x < 12.5). Draw a graph.
    [latex]\displaystyle{P}{({10}{<}{one thousand})}={(\text{base})}{(\text{height})}={({12.5}-{0})}{(\frac{{1}}{{15}})}={0.8333}\\[/latex]
    The probability a person waits less than 12.v minutes is 0.8333.
    This shows the graph of the function f(x) = 1/15. A horiztonal line ranges from the point (0, 1/15) to the point (15, 1/15). A vertical line extends from the x-axis to the end of the line at point (15, 1/15) creating a rectangle. A region is shaded inside the rectangle from x = 0 to x = 12.5.
  2. [latex]\displaystyle{\mu}=\frac{{{a}+{b}}}{{2}}=\frac{{{xv}+{0}}}{{2}}={7.5}\\[/latex]. On the average, a person must await 7.v minutes.[latex]\displaystyle{\sigma}=\sqrt{{\frac{{{({b}-{a})}^{{ii}}}}{{12}}}}=\sqrt{{\frac{{{({xv}-{0})}^{{2}}}}{{12}}}}={4.3}\\[/latex]The standard divergence is 4.3 minutes.
  3. Observe the 90th percentile. Draw a graph. Permit m = the 90th percentile.
    [latex]\displaystyle{P}(x<yard)=(\text{base of operations})(\text{meridian})=(k-0)(\frac{ane}{fifteen})\\[/latex]
    [latex]\displaystyle{0.90}=(k)(\frac{one}{15})\\[/latex]
    [latex]g=(0.90)(15)=13.five\\[/latex]
    k is sometimes called a critical value.The 90th percentile is 13.5 minutes. Ninety percentage of the time, a person must wait at well-nigh 13.5 minutes.
    f(X)=1/15 graph displaying a boxed region consisting of a horizontal line extending to the right from point 1/15 on the y-axis, a vertical upward line from an arbitrary point on the x-axis, and the x and y-axes. A shaded region from points 0-k occurs within this area. The area of this probability region is equal to 0.90.

Try It

The total elapsing of baseball game games in the major league in the 2022 season is uniformly distributed between 447 hours and 521 hours inclusive.

  1. Find a and b and describe what they correspond.
  2. Write the distribution.
  3. Observe the hateful and the standard difference.
  4. What is the probability that the elapsing of games for a team for the 2022 season is between 480 and 500 hours?
  5. What is the 65th percentile for the duration of games for a team for the 2022 season?
  1. a is 447, and b is 521. a is the minimum duration of games for a team for the 2022 flavor, and b is the maximum elapsing of games for a team for the 2022 flavor.
  2. X ~ U (447, 521).
  3. μ = 484, and σ = 21.36
  4. P(480 < 10 < 500) = 0.2703
  5. 65th percentile is 495.i hours.

Example 4

Suppose the time it takes a nine-year sometime to swallow a donut is betwixt 0.v and 4 minutes, inclusive. PermitTen = the time, in minutes, it takes a nine-twelvemonth old kid to consume a donut. Then X~ U (0.v, 4).

  1. The probability that a randomly selected nine-yr erstwhile child eats a donut in at least two minutes is _______.
  2. Observe the probability that a dissimilar nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.five minutes.

Solution

  1. 0.5714
  2. This question has a conditional probability. You are asked to observe the probability that a ix-year old kid eats a donut in more than two minutes given that the child has already been eating the donut for more 1.5 minutes. Solve the problem two different ways (see Example 3). You must reduce the sample space.
    1. First way: Since yous know the child has already been eating the donut for more than 1.v minutes, yous are no longer starting at a = 0.5 minutes. Your starting bespeak is one.5 minutes. Write a newf(x):[latex]\displaystyle{f{{({x})}}}=\frac{{i}}{{{4}-{1.5}}}={25} \text{ for } {1.five}\leq{ten}\leq{4}\\[/latex]FindP(ten > 2|x > one.5). Describe a graph.
      f(X)=2/5 graph displaying a boxed region consisting of a horizontal line extending to the right from point 2/5 on the y-axis, a vertical upward line from points 1.5 and 4 on the x-axis, and the x-axis. A shaded region from points 2-4 occurs within this area. [latex]\displaystyle{P}{({x}{>}{2}|{x}{>}{one.5})}={(\text{base of operations})}{(\text{new height})}={({4}-{2})}{(\frac{{2}}{{5}})}=\frac{{4}}{{v}}\\[/latex]
      The probability that a nine-twelvemonth old child eats a donut in more than two minutes given that the child has already been eating the donut for more than than 1.5 minutes is [latex]\displaystyle\frac{{4}}{{5}}\\[/latex].
    2. Second way: Depict the original graph for X ~ U (0.5, 4). Use the conditional formula[latex]\displaystyle{P}{({x}{>}{two}{mid}{ten}{>}{1.5})}=\frac{{{P}{({x}{>}{ii} \text{ AND } {x}{>}{one.v})}}}{{{P}{({ten}{>}{i.5})}}}=\frac{{{P}{({10}{>}{2})}}}{{{P}{({10}{>}{i.5})}}}=\frac{{\frac{{2}}{{3.v}}}}{{\frac{{two.five}}{{3.v}}}}={0.8}=\frac{{four}}{{5}}\\[/latex]

Try It

Suppose the time it takes a educatee to end a quiz is uniformly distributed between six and 15 minutes, inclusive. Allow10 = the fourth dimension, in minutes, information technology takes a educatee to cease a quiz. So Ten ~ U (6, 15).

Detect the probability that a randomly selected educatee needs at to the lowest degree eight minutes to consummate the quiz. Then detect the probability that a different pupil needs at least eight minutes to finish the quiz given that she has already taken more than vii minutes.

P (x > 8) = 0.7778

P (x > eight | x > seven) = 0.875

Example 5

Ace Heating and Ac Service finds that the amount of time a repairman needs to prepare a furnace is uniformly distributed between 1.v and four hours. Letx = the time needed to fix a furnace. So 10 ~ U (one.5, 4).

  1. Detect the probability that a randomly selected furnace repair requires more than two hours.
  2. Discover the probability that a randomly selected furnace repair requires less than three hours.
  3. Find the 30th percentile of furnace repair times.
  4. The longest 25% of furnace repair times take at least how long? (In other words: find the minimum time for the longest 25% of repair times.) What percentile does this represent?
  5. Find the mean and standard deviation

Solution

  1. To detectf(10): [latex]\displaystyle{f{{({x})}}}=\frac{{1}}{{{4}-{1.five}}}={12.5} \text{ and so } {f{{({x})}}}={0.four}\\[/latex]
    P(10 > two) = (base of operations)(peak) = (4 – 2)(0.four) = 0.8
    This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 2 to x = 4.
    Compatible Distribution between 1.v and 4 with shaded area between two and four representing the probability that the repair fourth dimension
    10 is greater than two
  2. P(x < 3) = (base of operations)(height) = (3 – i.5)(0.4) = 0.6The graph of the rectangle showing the entire distribution would remain the aforementioned. Still the graph should exist shaded between
    x = one.5 and x = iii. Annotation that the shaded area starts at ten = one.five rather than at x = 0; since X ~ U (one.5, 4), x tin not exist less than 1.v.
    This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the point (1.5, 0.4) to the point (4, 0.4). Vertical lines extend from the x-axis to the graph at x = 1.5 and x = 4 creating a rectangle. A region is shaded inside the rectangle from x = 1.5 to x = 3.Uniform Distribution between 1.v and four with shaded area between one.5 and 3 representing the probability that the repair timex is less than 3
  3. This shows the graph of the function f(x) = 0.4. A horiztonal line ranges from the indicate (1.5, 0.4) to the point (4, 0.four). Vertical lines extend from the ten-axis to the graph at ten = 1.5 and x = iv creating a rectangle. A region is shaded inside the rectangle from x = i.5 to ten = 1000. The shaded expanse represents P(ten < k) = 0.3.
    Uniform Distribution between 1.v and iv with an area of 0.30 shaded to the left, representing the shortest thirty% of repair times.
    P (x < k) = 0.thirty
    P(ten < 1000) = (base of operations)(superlative) = (k – one.5)(0.4)0.three = (chiliad – 1.5) (0.4); Solve to find k:0.75 =yard – 1.5, obtained by dividing both sides past 0.4
    k = two.25 , obtained by adding 1.5 to both sidesThe 30th percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or less.

  4. Uniform Distribution between one.5 and four with an area of 0.25 shaded to the right representing the longest 25% of repair times.P(x > k) = 0.25
    P(x > k) = (base)(peak) = (4 – thousand)(0.4)0.25 = (four –k)(0.4); Solve for yard:0.625 = iv −k, obtained by dividing both sides by 0.4

    −3.375 = −thou, obtained by subtracting four from both sides: k = 3.375

    The longest 25% of furnace repairs have at least 3.375 hours (3.375 hours or longer).

    Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are iii.375 hours or less. 3.375 hours is the 75th percentile of furnace repair times.

  5. [latex]{\mu}={\frac{a+b}{2}}\text{ and }{\sigma}=\sqrt{\frac{(b-a)^2}{12}}\\[/latex]
    [latex]{\mu}=\frac{1.v+4}{2}=2.75\text{ hours and }{\sigma}=\sqrt{\frac{(4-1.5)^ii}{12}}= 0.7217 \text{ hours}\\[/latex]

Try It

The corporeality of time a service technician needs to change the oil in a car is uniformly distributed betwixt 11 and 21 minutes. AllowX = the time needed to modify the oil on a car.

  1. Write the random variable X in words. Ten = __________________.
  2. Write the distribution.
  3. Graph the distribution.
  4. Observe P (x > nineteen).
  5. Find the 50th percentile.
  1. Permit X = the time needed to modify the oil in a auto.
  2. Ten ~ U (11, 21).
  3. This graph shows a uniform distribution. The horizontal axis ranges from 405 to 525. The distribution is modeled by a rectangle extending from x = 447 to x = 521.
  4. P (10 > 19) = 0.2
  5. the 50th percentile is 16 minutes.

References

McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, 1995.

Concept Review

IfTen has a uniform distribution where a < x < b or a ≤ ten ≤ b, and then 10 takes on values between a and b (may include a and b). All values x are every bit likely. We write X ∼ U(a, b). The mean of X is [latex]\displaystyle{\mu}=\frac{{{a}+{b}}}{{two}}\\[/latex]. X is continuous.

The graph shows a rectangle with total area equal to 1. The rectangle extends from x = a to x = b on the x-axis and has a height of 1/(b-a).

The probabilityP(c < Ten < d) may be found past calculating the surface area under f(ten), between c and d. Since the corresponding area is a rectangle, the area may exist establish but by multiplying the width and the height.

Formula Review

X = a real number betwixt a and b (in some instances, X can take on the values a and b). a = smallest X; b = largest 10

10 ~ U (a, b)

The hateful is [latex]\displaystyle\mu=\frac{{{a}+{b}}}{{2}}\\[/latex]

The standard departure is [latex]\displaystyle\sigma=\sqrt{{\frac{{({b}-{a})}^{{ii}}}{{12}}}}\\[/latex]

Probability density function: [latex]\displaystyle{f{{({x})}}}=\frac{{one}}{{{b}-{a}}} \text{ for } {a}\leq{X}\leq{b}\\[/latex]

Area to the Left of10: [latex]\displaystyle{P}{({X}{<}{10})}={({x}-{a})}{(\frac{{1}}{{{b}-{a}}})}\\[/latex]

Expanse to the Right ofx: [latex]\displaystyle{P}{({Ten}{>}{x})}={({b}-{x})}{(\frac{{ane}}{{{b}-{a}}})}\\[/latex]

Area Betweenc and d: [latex]\displaystyle{P}{({c}{<}{x}{<}{d})}={(\text{base of operations})}{(\text{height})}={({d}-{c})}{(\frac{{1}}{{{b}-{a}}})}\\[/latex]

Uniform:X ~ U(a, b) where a < x < b

  • pdf: [latex]\displaystyle{f{{({x})}}}=\frac{{1}}{{{b}-{a}}}\\[/latex] for a ≤ x ≤ b
  • cdf: P(X ≤ x) = [latex]\displaystyle\frac{{{x}-{a}}}{{{b}-{a}}}\\[/latex]
  • mean: [latex]\displaystyle\mu=\frac{{{a}+{b}}}{{ii}}\\[/latex]
  • standard deviation: [latex]\displaystyle\sigma=\sqrt{{\frac{{({b}-{a})}^{{2}}}{{12}}}}\\[/latex]
  • P(c < X < d) = (d – c)

What Is The Population Mean Of The Times To Repair?,

Source: https://courses.lumenlearning.com/odessa-introstats1-1/chapter/the-uniform-distribution/

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